given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

Correct so far. Since the equation you quote is quadratic, it has two solutions. The first is obviously t=0 (it starts off at ground level). This leaves

given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

You could make use of the fact that the projectile come down just as fast as it went up, so final velocity is -14 m/s

That way you an use the non-quadratic formula v = u + at [not your symbols but a common set]

given the initial velocity, it is acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is Voy = +14 m/s, therefore we have y = 0m ay = -9.8m/s^2 V0y = +14m/s t = ?

The time of flight can be determined by the equation y = voyt + 1/2ayt^2

You don't need to use the quadratic equation.
I offer a more pictorial method for this problem.

There are two important times during the parabola trajectory, tA and tB.
tA is when the object is at the peak of the parabola.
tB is when the object lands.